Mistake Master

Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms AB & BC

L'Hospital's Rule replaces $\lim \frac{f}{g}$ with $\lim \frac{f'}{g'}$, but only after substitution has confirmed the form is $\frac{0}{0}$ or $\frac{\infty}{\infty}$. It differentiates the numerator and denominator separately, which is not the quotient rule, and it may be applied again as long as the form still qualifies, so $\frac{1-\cos x}{x^2}$ takes two rounds to reach $\frac{1}{2}$. Products, differences, and exponential forms have to be rewritten as quotients first.

Two failures dominate. The rule is used without checking, so a determinate form such as $\frac{0}{1}$ gets differentiated and returns a confident wrong number, or the quotient rule is applied by mistake, or one application is treated as enough while the form is still indeterminate. And the result is misread: the number the rule produces is the limit of the original quotient, not a value of the function at the point and not the derivative of the quotient.

STEP 0, ALWAYS: SUBSTITUTE AND READ THE FORM what is lim f/g at c? 0/0 or ∞/∞ 5/7, 0/5 5/0 RULE APPLIES that IS the limit infinite or one-sided go to f′/g′, then recheck nothing to resolve the rule says nothing NEED REWRITING FIRST, into a quotient: 0 · ∞ move a factor down ∞ - ∞ common denominator 1^∞, 0⁰, ∞⁰ take a log lim (x²-4)/(x-1) at x = 2 is 0/1 = 0. Differentiating anyway returns 4, not the limit.
Only the left branch permits the rule. Skipping the substitution step is what routes a determinate form into it.
lim (eˣ - 1)/x AS x → 0, WHICH IS 0/0, SO THE RULE IS AVAILABLE L’HOSPITAL: TOP AND BOTTOM THE QUOTIENT RULE, BY MISTAKE f = eˣ - 1 f′ = eˣ (f/g)′ = (f′g - fg′)/g² g = x g′ = 1 = (xeˣ - (eˣ - 1)) / x² f′/g′ = eˣ/1 → 1 → 0/0 again, and it is a different function entirely two separate derivatives WHY it works: near c both pieces are their own linearizations, so the ratio is f′(c)(x - c) over g′(c)(x - c), and the equal steps cancel. Topic 4.6 stands behind 4.7.
Two rules that look alike and share nothing. One takes two derivatives; the other takes one, of a different object.

The work

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Lesson
Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms

Treats the indeterminate form as a precondition to be verified rather than assumed, differentiates numerator and denominator separately instead of as a quotient, repeats the rule while the form still qualifies, rewrites products and differences into quotients, and reads the result as a limit of the original expression.

Skill check · 10 scenarios
Diagnostic
10-item topic check

Ten items spanning the two failure modes of this topic: applying L'Hospital's Rule to a form that is not indeterminate, or as though it were the quotient rule, or stopping while the form still qualifies, and misreading the resulting number as a function value or a derivative.

Not yet available · 10 items
Targeted Practice
Drill a single misconception

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears the misconception and moves you to the next.

Take the diagnostic to identify your misconceptions